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250/500 V transformer: max-efficiency output

Short answer

Shunt branch: Ro=781.3 Ω, Xo=263.9 Ω; series referred to primary: R01=0.174 Ω, X01=0.379 Ω. For maximum efficiency at 0.9 pf lag, output power is 4.83 kW (efficiency ≈ 96.8%).

Test data on a 1-Ø, 250/500 V, 50 Hz transformer are: O.C. Test: 250 V, 1 A, 80 W (carried on L.V. side) S.C. Test: 20 V, 12 A, 100 W (carried on H.V. side) Then draw the equivalent circuit referred to primary side and find out the output power to obtain maximum efficiency at 0.9 lag p.f.

Given: 1-φ, 250/500 V, 50 Hz. OC (LV): 250 V, 1 A, 80 W. SC (HV): 20 V, 12 A, 100 W. Primary = 250 V side; K = 2.

Assumption: the rated kVA is not given. The SC test current of 12 A on the HV side is taken as rated, so S = 500 × 12 = 6 kVA.

Shunt branch (primary side)

cos φ0 = 80/(250 × 1) = 0.32
Iw = 0.32 A ; Iμ = √(1 − 0.32²) = 0.947 A
Ro = 250/0.32 = 781.3 Ω ; Xo = 250/0.947 = 263.9 Ω

Series branch

On the HV side: Z02 = 20/12 = 1.667 Ω; R02 = 100/12² = 0.694 Ω; X02 = √(1.667² − 0.694²) = 1.515 Ω. Referred to the primary (÷ 4):

Z01 = 0.417 Ω ; R01 = 0.1736 Ω ; X01 = 0.379 Ω

Equivalent circuit (primary side)

        R01=0.174Ω  X01=0.379Ω
 o──┬────┬───/\/\/───────(((────────o
    │    │                          
 781.3Ω 263.9Ω                  load
    │    │                       V2'
 o──┴────┴───────────────────────────o

Output for maximum efficiency at 0.9 lagging

Iron loss Pi = 80 W ; full-load Cu loss Pcu = 100 W
x = √(Pi/Pcu) = √(80/100) = 0.894
Load kVA at max η = 0.894 × 6 = 5.37 kVA
Output power = 5.37 × 0.9 = 4.83 kW
η_max = 4830/(4830 + 2 × 80) = 96.8 %

Answer: output power ≈ 4.83 kW (0.894 of full load) for maximum efficiency; η_max ≈ 96.8 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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